Generalized Calculus III Summer 2015, Session II Thursday, July 23, 2015
Agenda 1. 2. 3. 4.
Motivation Defective matrices cannot be diagonalized because they do not possess enough eigenvectors to make a basis. How can we correct this defect? Example [ ] 1 1 The matrix A = is defective. 0 1 1. Only eigenvalue is λ = 1. [ ] 0 1 2. A I = 0 0 3. Single eigenvector v = (1, 0). 4. We could use u = (0, 1) to complete a basis. 5. Notice that (A I)u = v and (A I) 2 u = 0. Maybe we just didn t multiply by A λi enough times.
If A is an n n matrix, a generalized eigenvector of A corresponding to the eigenvalue λ is a nonzero vector x satisfying (A λi) p x = 0 for some positive integer p. Equivalently, it is a nonzero element of the nullspace of (A λi) p. Example are generalized eigenvectors with p = 1. In the previous example we saw that v = (1, 0) and u = (0, 1) are generalized eigenvectors for [ ] 1 1 A = and λ = 1. 0 1
Computing generalized eigenvectors Example Determine generalized eigenvectors for the matrix 1 1 0 A = 0 1 2. 0 0 3 1. Characteristic polynomial is (3 λ)(1 λ) 2. 2. Eigenvalues are λ = 1, 3. 3. are λ 1 = 3 : v 1 = (1, 2, 2), λ 2 = 1 : v 2 = (1, 0, 0). 4. Final generalized eigenvector will a vector v 3 0 such that (A λ 2 I) 2 v 3 = 0 but (A λ 2 I) v 3 0. Pick v 3 = (0, 1, 0). Note that (A λ 2 I)v 3 = v 2.
Facts about generalized eigenvectors How many powers of (A λi) do we need to compute in order to find all of the generalized eigenvectors for λ? Fact If A is an n n matrix and λ is an eigenvalue with algebraic multiplicity k, then the set of generalized eigenvectors for λ consists of the nonzero elements of nullspace ( (A λi) k). In other words, we need to take at most k powers of A λi to find all of the generalized eigenvectors for λ.
Computing generalized eigenvectors Example Determine generalized eigenvectors for the matrix 1 2 0 A = 1 1 2. 0 1 1 1. Single eigenvalue of λ = 1. 2. Single eigenvector v 1 = ( 2, 0, 1). 3. Look at 2 0 4 (A I) 2 = 0 0 0 1 0 2 to find generalized eigenvector v 2 = (0, 1, 0). 4. Finally, (A I) 3 = 0, so we get v 3 = (1, 0, 0).
Facts about generalized eigenvectors The aim of generalized eigenvectors was to enlarge a set of linearly independent eigenvectors to make a basis. Are there always enough generalized eigenvectors to do so? Fact If λ is an eigenvalue of A with algebraic multiplicity k, then ( nullity (A λi) k) = k. In other words, there are k linearly independent generalized eigenvectors for λ. Corollary If A is an n n matrix, then there is a basis for R n consisting of generalized eigenvectors of A.
Computing generalized eigenvectors Example Determine generalized eigenvectors for the matrix 1 2 0 A = 1 1 2. 0 1 1 1. From last time, we have eigenvalue λ = 1 and eigenvector v 1 = ( 2, 0, 1). 2. Solve (A I)v 2 = v 1 to get v 2 = (0, 1, 0). 3. Solve (A I)v 3 = v 2 to get v 3 = ( 1, 0, 0).
of generalized eigenvectors Let A be an n n matrix and v a generalized eigenvector of A corresponding to the eigenvalue λ. This means that for a positive integer p. If 0 q < p, then (A λi) p v = 0 (A λi) p q (A λi) q v = 0. That is, (A λi) q v is also a generalized eigenvector corresponding to λ for q = 0, 1,..., p 1. If p is the smallest positive integer such that (A λi) p v = 0, then the sequence (A λi) p 1 v, (A λi) p 2 v,..., (A λi) v, v is called a chain or cycle of generalized eigenvectors. The integer p is called the length of the cycle.
of generalized eigenvectors Example In the previous example, 0 2 0 A λi = 1 0 2 0 1 0 and we found the chain 1 0 2 v = 0, (A λi)v = 1, (A λi) 2 v = 0. 0 0 1 Remark The terminal vector in a chain is always an eigenvector. Fact The generalized eigenvectors in a chain are linearly independent.
Introduction to form What s the analogue of diagonalization for defective matrices? That is, if {v 1, v 2,..., v n } are the linearly independent generalized eigenvectors of A occurring in chains, what does the matrix S 1 AS look like, where S = [ v 1 v 2 v n ]? Suppose that v 1, v 1,..., v k is a chain of generalized eigenvectors, so that (A λi)v i = v i 1 for i > 1 and (A λi)v 1 = 0. Then we have Av i = λv i + v i 1 for i > 1 and Av 1 = λv 1.
blocks The matrix for T (x) = Ax with respect to a basis consisting of a chain of generalized eigenvectors will be a block: If λ is a real number, then the square matrix of the form λ 1 0 0 0 0 λ 1 0 0 0 0 λ 1 0 J λ =.......... 0 0 λ 1 0 0 0 λ is called a block corresponding to λ.
In general, we will need to find more than one chain of generalized eigenvectors in order to have enough for a basis. Each chain will be represented by a block. A square matrix consisting of blocks centered along the main diagonal and zeros elsewhere is said to be in (JCF). Theorem If S is the matrix whose columns are a basis of generalized eigenvectors of A arranged in chains, then S 1 AS is a matrix in JCF. It is unique up to a rearrangement of the blocks. We may therefore refer to this matrix as the canonical form of A, and we see that every matrix is similar to a matrix in JCF.
Examples The matrix 2 1 0 0 2 1 0 0 2 5 1 0 5 7 1 0 7 is in JCF. It contains five blocks. Any diagonal matrix is in JCF. All of its blocks are 1 1. The matrix [ 0 ] 1 0 0 0 0 0 0 0 7 9 is in JCF. It has two blocks of sizes 2 and 1.
Uses of JCF Theorem Two n n matrices are similar if and only if they have the same (up to a rearrangement of the blocks). Our main use for JCF will be solving x = Ax when the matrix A is defective.